Hi!

How do i proof that 9^(n+1) == 8n+9 (mod 64) congruence has infinite solutions if n is a natural number?

Thanks!

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- Nov 8th 2009, 08:48 AMBen92non-linear congruence proof
Hi!

How do i proof that 9^(n+1) == 8n+9 (mod 64) congruence has infinite solutions if n is a natural number?

Thanks! - Nov 8th 2009, 10:21 AMDrexel28
Prove that the congruence $\displaystyle 9^{n+1}\equiv8n+9\text{ mod }64$ has infinite solutions in the natural numbers.**Problem:**

**Proof:**

**Lemma:**$\displaystyle 9^{64n}\equiv 1\text{ mod }64\quad\forall n\in\mathbb{N}$

**Proof:**We do this by weak induction.

__Base case:__Merely note that $\displaystyle 9^{64}-1=\left(9^{32}-1\right)\left(9^{32}+1\right)=\cdots=\left(9^1-1\right)\cdot(9^1+1)\cdot(9^2+1)\cdots\left(9^{32} +1\right)$. Now noting that the first term is $\displaystyle 9-1=8$ and that we are multiplying by 5 even numbers gives the desired result.

__Induction hypothesis:__Assume that $\displaystyle 9^{64n}\equiv1\text{ mod }64$

__Inductive step:__Merely noting that $\displaystyle 9^{64(n+1)}=9^{64n}\cdot9^{64}$ and using the base case and inductive hypothesis in conjunction finishes the proof.

Now using the above lemma we can clearly see that $\displaystyle n=64m$ is a solution for $\displaystyle m\in\mathbb{N}$ because $\displaystyle 9^{64m+1}=9^{64m}\cdot9^1\equiv9\equiv0+9\equiv8\c dot64m+9$. And since there are clearly infinite number of natural numbers of the form $\displaystyle 64m$ this provides the desired result.

*Remark:*Choosing $\displaystyle n=32m$ would work equally as well.

EDIT: I feel as though I should explain my "remark" better. The operative part of this proof is that we needed $\displaystyle 64|9^{64}-1$. You'll notice that this was possible because when we factored $\displaystyle 9^{64}-1$ we were left with something of the form $\displaystyle 8\cdot e \cdot e \cdot e \cdot e \cdot e$ where $\displaystyle e$ is an even number. Although it wasn't mentioned we only really needed the first three $\displaystyle e$s, the rest were just extra. So in fact we could have gotten by with just 3 or 4 $\displaystyle e$s and in fact any number which factored yields $\displaystyle 8\underbrace{e\cdot e\cdot e\cdots}_{\ge3}$ would suffice. For this reason we can actually say that $\displaystyle n=2^xm$ is sufficent for $\displaystyle x\ge4$. Also, there is an even "easier" and more general way to do this. I hope that you take the time to explore it. - Nov 8th 2009, 12:52 PMBruno J.
A bit shorter : by Euler's theorem, $\displaystyle a^{k\phi(m)}\equiv 1 \mod m$ for any $\displaystyle a$ relatively prime to $\displaystyle m$, and for all $\displaystyle k \in \mathbb{Z}$. Since $\displaystyle \phi(64)=32$, and $\displaystyle (9,64)=1$, we have $\displaystyle 9^{32k} \equiv 1 \mod 64$. So $\displaystyle 9^{32k+1}\equiv 9 \equiv 9+8\times 32k \mod 64$.

- Nov 8th 2009, 01:05 PMBen92
Thanks Drexel and Bruno.